Logarithm Test 1

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Logarithm Test 1

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Question 1/10
1 / -0

If log (x⁵ y²) = 5a + 2b and log (x² y⁵) = 2a + 5b, find log (xy) in terms of a and b?

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A
a + b

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B
ab

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C
a² +b²

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D
a/b

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Solutions

log (x⁵ y²) = 5a + 2b

⇒ 5 logx + 2logy = 5a + 2b → (1)

Similarly, log (x² y⁵) = 2a + 5b

⇒ 2 log x+ 5 log y = 2a + 5b → (2)

∴ log (xy) = log x + log y

Adding (1) and (2) 7 log x + 7 log y = 7a + 7b

log x + log y = a + b
Question 2/10
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fraction numerator 1 over denominator log subscript 2 straight x end fraction plus fraction numerator 1 over denominator log subscript 2 straight x end fraction plus fraction numerator 1 over denominator log subscript 4 straight x end fraction plus fraction numerator 1 over denominator log subscript 5 straight x end fraction plus fraction numerator 1 over denominator log subscript 6 straight x end fraction plus fraction numerator 1 over denominator log subscript 7 straight x end fraction

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A
1 / log_27^x

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B
log_27^x

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C
log₅₀₄₀x

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D
log_x5040

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Solutions

Similarly,

∴

=

= = log_x 5040
Question 3/10
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If log_x 4 = 0.4, then the value of x is:

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A
1

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B
32

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C
16

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D
4

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Solutions

log_x 4 = 0.4 ⇔ log_x 4 = ⇔x²^/5= 4 ⇔ x = 4^5/2= (2²)^5/2

⇔ x = = 2⁵ ⇔ x = 32.
Question 4/10
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The value of log₂ [log₂ log₂ log₂ (65536)] is

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A
8

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B
16

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C
4

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D
1

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Solutions

log₂ log₂ log₂ log₂ 2¹⁶ = log₂ log₂ log₂ (16)

= log₂ log₂ log₂ (2⁴) = log₂ log₂ (4)

= log₂ log₂ (2²) = log₂ (2) = 1
Question 5/10
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If the logarithm of 19683 to a base is 6, then what is the base?

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A
√3

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B
3√3

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C
9

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D
3

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Solutions

Let the base be x

Log_x 19683 = 6 ⇒ x⁶ = 19683 = (3)⁶

∴ x = 3
Question 6/10
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If 5 + log₁₀x = 5log₁₀y, then express x in terms of y.

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A
x = 5y⁵

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B
x = (y)

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C
x² = 10y⁵

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D
x = (y/, 10) ⁵

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Solutions

5 + log₁₀x = 5 log₁₀y

5 log₁₀10 + log₁₀x = log₁₀y⁵ ⇒ log₁₀x = log₁₀y⁵– log₁₀10⁵

⇒ log₁₀x = log₁₀ ⇒ x = y⁵/10⁵ = (y/10)⁵
Question 7/10
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The value of log₅is:

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A
3

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B
–3

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C
1/3

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D
-1/3

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Solutions

Let log₅ = n. Then, 5ⁿ = ⇒5ⁿ = 5^–3 ⇒ n = – 3.

∴ log₅ = –3.
Question 8/10
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Find the logarithm of 1728 to the base 2 square root of 3 .

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A
3.124

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B
3.1732

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C
6

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D
5

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Solutions

let 1728 = x ⇒ (2)^x = 1728

∵ 1728 = 2⁶ ()⁶ = (2)⁶

(2)^x = (2)⁶

On comparing ⇒ x = 6.
Question 9/10
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Find the logarithm of 64 × cube root of 512 to the base cube root of 8

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A
1

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B
3

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C
9

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D
12

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Solutions

=

= (3) (3) log₈8 = 9(1) = 9
Question 10/10
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Simplify: log₂ log₂ log₂log subscript square root of 3 end subscript 6561.

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A
0

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B
1

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C
2 log 2

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D
4 log 2

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Solutions

6151 = = 16

log₂ 16 = log₂2⁴ = 4

log₂4 = log₂2² = 2

log₂2 = 1

∴ log₂ log₂ log₂ log₂ 6561 = 1.

Question 1/10

a + b

ab

a2 +b2

a/b

Question 2/10

1 / log27x

log27x

log5040x

logx5040

Question 3/10

1

32

16

4

Question 4/10

8

16

4

1

Question 5/10

√3

3√3

9

3

Question 6/10

x = 5y5

x = (y)

x2 = 10y5

x = (y/, 10) 5

Question 7/10

3

–3

1/3

-1/3

Question 8/10

3.124

3.1732

6

5

Question 9/10

1

3

9

12

Question 10/10

0

1

2 log 2

4 log 2

a² +b²

1 / log_27^x

log_27^x

log₅₀₄₀x

log_x5040

x = 5y⁵

x² = 10y⁵

x = (y/, 10) ⁵